Attempt to solve f(y) = -5y^3+6y^2-4 = 0
First divide through by -y^3 to get:
5-6/y+4/y^3 = 0
Let x = 1/y
Then 4x^3-6x+5 = 0
Now let x = u + v
0 = 4(u+v)^3 - 6(u+v) + 5
=4u^3+4v^3+(12uv-6)(u+v)+5
=4u^3+4v^3+6(2uv-1)(u+v)+5
Let v = 1/(2u)
=4u^3+1/(2u^3)+5
Multiply through by 2u^3 to get:
8(u^3)^2+10(u^3)+1 = 0
u^3 = (-10+-sqrt(100-32))/16
=(-10+-sqrt(68))/16
=(-5+-sqrt(17))/8
Write:
u_1 = root(3)((-5+sqrt(17))/8)
v_1 = root(3)((-5-sqrt(17))/8)
Then real root of 4x^3-6x+5 = 0 is
x = u_1 + v_1
The other two (complex) roots are:
x = omega u_1 + omega^2 v_1
x = omega^2 u_1 + omega v_1
where omega = -1/2+sqrt(3)/2i
y = 1/x
So the real root of f(y) = 0 is y_1 = 1/(u_1+v_1)
and the complex roots are:
y_2 = 1/(omega u_1 + omega^2 v_1)
y_3 = 1/(omega^2 u_1 + omega v_1)
f(y) = -5(y - y_1)(y - y_2)(y - y_3)
